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VI Dynamics

25

A magic electromagnetic ¬eld

DONALD LYNDEN-BELL

Institute of Astronomy, The Observatories,

Madingley Road, Cambridge CB3 0HA, UK

and Clare College, Cambridge, UK

An electromagnetic ¬eld of simple algebraic structure is simply derived. It

turns out to be the G = 0 limit of the charged rotating Kerr-Newman metrics.

These all have gyromagnetic ratio 2, the same as the Dirac electron. The

charge and current distributions giving this high gyromagnetic ratio have

charges of both signs rotating at close to the velocity of light.

It is conjectured that something similar may occur in the quantum elec-

trodynamic charge distribution surrounding the point electron.

25.1 The electromagnetic ¬eld

Away from charges and currents, both the electrostatic potential, ¦, and

the magnetostatic potential, χ, are harmonic. Thus Ψ = ¦ + iχ satis¬es

∇2 Ψ = 0 .

The solution obeying this equation everywhere “ except the origin “ and

tending to zero at in¬nity is Ψ = q/r, but if we move the origin to b this

solution becomes

(r ’ b)2 .

Ψ=q

This solution is harmonic whether q and b are real or complex.

To ensure no magnetic monopole, term q must be real, but we now con-

sider the possibility that b = ia where a is real so that b is pure imag-

inary. Then we shall have both an electric and a magnetic ¬eld with

F = E + iB = ’∇Ψ. Without loss of generality we may orient the z

axis along a so that

1/2

R2 + (z ’ ia)2 R2 = x2 + y 2 .

Ψ=q where (25.1)

This expression will be harmonic except at singularities and branch points.

369

Lynden-Bell

370

The singularities lie at R = a and z = 0. If we ask for no branch points at

in¬nity then we may take the cut de¬ned by the disk z = 0, R ¤ a, (but

notice that we could take the cut around the sphere r = a, z ≥ 0 say).

We may evaluate ’∇Ψ to obtain

3/2

F = E + iB = q(r ’ ia) (r ’ ia)2 . (25.2)

The total charge is clearly q but the ¬eld also has a magnetic dipole moment.

Indeed for r > a we may use the Legendre polynomial expansion of Ψ

∞ n

q ia

Ψ= Pn (cos θ) . (25.3)

r r

0

Evidently all the P2n have real coe¬cients and all the P2n+1 have imaginary

coe¬cients so the magnetic potential is antisymmetrical about z = 0 while

the electric potential is symmetrical. Evidently the magnetic moment is the

coe¬cient of iP1 which is qa while the electric quadrupole moment is qa2 ,

etc. The relativistic invariants of the ¬eld are contained in

2

F 2 = E 2 ’ B 2 + 2iE · B = q 2 (r ’ ia)2 .

1

2

is only imaginary if (r ’ ia)2 = ± √ (1 ± i)|r ’ ia|2 which

Now (r ’ ia)2

2

occurs when r ’ a /(2r · a) = ±1 as then the real and imaginary parts

2 2

are equal in magnitude. This condition may be rewritten (r ± a)2 = 2a2

√

so E 2 = B 2 only on two spheres of radius 2 a centred on (r = ±a). The

circle in which they meet is the ring z = 0, r = a.

Figure 25.1 illustrates where |B| > |E|, etc. E and B are perpendicular

when (r ’ ia)2 = r 2 ’ a2 ’ 2ia · r is either purely real or purely imaginary;

i.e., on the sphere r = a, and the plane z = 0. The Poynting vector is given

by

F— — F = (E ’ iB) — (E + iB) = 2iE — B = 2iq 2 a — r

3

r 2 + a2 ,

and the ¬eld energy density by (8π)’1 F— · F = (8π)’1 E 2 + B 2 . The

velocity of the Lorentz frame in which E and B are parallel is given by

v = cV where

= E—B E2 + B2 = a — r

1+V2 a2 + r 2

V

= F— — F /(2iF · F— ) ;

A magic electromagnetic ¬eld 371

squaring and solving for V we ¬nd

(a2 + r 2 )2 ’ 4a2 R

a2 + r 2 ’

V = /(2aR)

(a2 + r 2 )2 ’ 4a2 R2

a2 + r 2 + a2 + »

= 2aR = aR ,

where

(r 2 ’ a2 )2 + 4(a · r)2

r 2 ’ a2 +

1

»= ,

2

is de¬ned with the positive root and µ is the same but for the negative root.

» and µ are spheroidal coordinates. Evidently „¦ = V /R is constant on the

confocal spheroids, » = constant which have a focal ring at the singularity.

(This result is due to J. Gair.)

E>B

E<B